Polymerisation

📚The chemistry

AS polymers were addition polymers: alkene monomers joining with no other product (topic 20). Condensation polymers are made from monomers with two functional groups each, which react to form ester or amide links, and a small molecule — water or HCl — is eliminated at every link.

35.1 Polyesters

Ester links, –COO–, form between –COOH (or –COCl) and –OH groups.

• A diol and a dicarboxylic acid: each end of each monomer forms an ester link, eliminating water. The standard example is PET (Terylene), from benzene-1,4-dicarboxylic acid and ethane-1,2-diol: \[ \mathrm{n\,HOOCC_6H_4COOH + n\,HOCH_2CH_2OH \rightarrow {-}\!\left[OCC_6H_4COOCH_2CH_2O\right]_n\!{-} + 2n\,H_2O} \] The repeat unit is –OC–C₆H₄–COO–CH₂CH₂–O–.
• A diol and a dioyl chloride: the same polymer forms faster, at room temperature, eliminating HCl instead of water.
• A hydroxycarboxylic acid: one monomer carrying both groups can link to itself. 2-hydroxypropanoic (lactic) acid gives poly(lactic acid), PLA: \[ \mathrm{n\,HOCH(CH_3)COOH \rightarrow {-}\!\left[OCH(CH_3)CO\right]_n\!{-} + n\,H_2O} \]

Polyamides

Amide links, –CONH–, form between –COOH (or –COCl) and –NH₂ groups.

• A diamine and a dicarboxylic acid: nylon-6,6 from hexane-1,6-diamine and hexanedioic acid (the numbers count the carbons in each monomer): \[ \mathrm{n\,H_2N(CH_2)_6NH_2 + n\,HOOC(CH_2)_4COOH \rightarrow {-}\!\left[NH(CH_2)_6NHCO(CH_2)_4CO\right]_n\!{-} + 2n\,H_2O} \]
• A diamine and a dioyl chloride (e.g. hexanedioyl dichloride): faster, eliminating HCl. Kevlar is made from benzene-1,4-diamine and benzene-1,4-dicarbonyl dichloride this way.
• An aminocarboxylic acid: 6-aminohexanoic acid polymerises with itself to nylon-6, repeat unit –NH(CH₂)₅CO–.
• Amino acids: joining by peptide bonds gives polypeptides and proteins (topic 34). A protein is a natural polyamide.

Repeat units and monomers

From monomers to repeat unit: remove the atoms lost in the condensation (OH from the acid and H from the alcohol or amine for water; Cl and H for HCl), join the ends with an ester or amide link, and draw one of each monomer’s residue in the bracket, with continuation bonds at both ends. For two different monomers, the repeat unit contains one of each.

From a polymer section back to monomers:

1. Find the ester (–COO–) or amide (–CONH–) links.
2. Break each link between the C=O carbon and the O or N.
3. Add OH to the C=O end (giving COOH, or Cl for a dioyl chloride) and H to the O or N end (giving OH or NH₂).

35.2 Addition or condensation?

• A monomer with a C=C double bond (and no second reactive group) → addition polymerisation. The polymer’s main chain is all carbon, and there is no other product.
• Monomers with two reactive functional groups (OH, COOH, COCl, NH₂) → condensation polymerisation. The main chain contains ester or amide links, and water or HCl is eliminated.

Looking at a polymer section: if the main chain is only carbon, it is an addition polymer; if it contains –COO– or –CONH–, it is a condensation polymer.

35.3 Degradable polymers

• Poly(alkene)s are chemically inert: their chains of strong, non-polar C–C and C–H bonds give nothing for water, acids, alkalis or enzymes to attack, so they are very difficult to biodegrade.
• Some polymers are photodegradable: they contain groups (such as C=O) that absorb ultraviolet light, and the absorbed energy breaks bonds in the chain, so the polymer breaks down in sunlight.
• Polyesters and polyamides can be hydrolysed: their ester and amide links are broken by acidic or alkaline hydrolysis (the reverse of condensation), giving back the monomers or their salts. That makes them biodegradable, given time — PLA, made from lactic acid from plant sugars, is used for compostable packaging for this reason.

✏️Worked example

(a) Draw the repeat unit of nylon-6,6 and calculate its M_r. [A_r: C 12.0, H 1.0, N 14.0, O 16.0] (b) This is a section of a polymer: –OCH₂CH₂OCO(CH₂)₄COOCH₂CH₂OCO(CH₂)₄CO–. State the type of polymer and identify its monomers. (c) Explain why nylon-6,6 can be broken down by hot aqueous acid but poly(propene) cannot.

(a) The repeat unit is –NH(CH₂)₆NHCO(CH₂)₄CO–, C₁₂H₂₂N₂O₂:

\[ M_\mathrm{r} = 12(12.0) + 22(1.0) + 2(14.0) + 2(16.0) = 226 \]

Check against the monomers: hexane-1,6-diamine (116) + hexanedioic acid (146) − 2 H₂O (36) = 226.

(b) The chain contains ester links, –COO–, so it is a polyester, made by condensation. Breaking each ester link and adding H to the O ends and OH to the C=O ends gives ethane-1,2-diol, HOCH₂CH₂OH, and hexanedioic acid, HOOC(CH₂)₄COOH.

(c) Nylon-6,6 contains amide links, which are hydrolysed by hot aqueous acid, breaking the chain back into hexanedioic acid and the salt of hexane-1,6-diamine. Poly(propene) is an addition polymer whose chain is made only of strong, non-polar C–C and C–H bonds, with no polar link for water to attack, so it is inert.

Check it. In a condensation polymer, the repeat unit mass equals the sum of the monomer masses minus the molecules eliminated per repeat unit (two waters here, one per link). If you get the plain sum of the monomers, the water has not been removed.

Putting only one monomer in the repeat unit of an AB-type polymer, or two waters in a self-condensing one. Nylon-6,6 needs one diamine and one diacid residue in the bracket. Nylon-6 and PLA, made from a single monomer with both groups, lose only one water per repeat unit. Count the new links formed per repeat unit to get the molecules eliminated.